** Mr. John Power entered his Trig class…. happy**

Mr. John Power entered his Trig class. He looked very happy. Then and there, some students sensed that Mr. Power was up to something. They knew that he liked to challenge them, and his being happy meant that he had come up with a hard one to crack.

Mr. Power addressed the class: Today, instead of a lesson you will have an interesting question to solve. For this, many students can work together. You are sitting in three rows. Each row can form a group. You have 20 minutes to solve the problem.

*The Problem: Determine the value of π using your knowledge of Trig*.

Most students were awed. Why would Mr. Power torture them like this? They did not have a clue on how to approach the problem because he had never talked about it in the class. All they remembered was definitions of some Trig functions and a few identities that they had memorized. It never came up in any of the worksheets they did. They just looked at each other wondering if someone else was going to come up with an answer.

After the 20 min passed, he asked the row 1 students who replied: Mr. Power, we have not yet covered this in class.

From row 2, one of the students proudly said: We know that cos inverse (also called arccos) of -1 is π radians. We used our calculators and determined the value of cos inverse of -1 and got a value of 3.14159265359. So π = 3.14159265359.

Mr. Power was frustrated because this was almost as bad as just getting the value of π from the Internet but said, “Good effort”.

Then he turned to the third group and asked Sara: Hope your group did better than that.

Sara said yes, and he called her to the board to explain what they did.

# Sara: Actually, it was all the genius of Tommy.

Tommy said that he was hungry and wished that he had some Hawaiian pizza. That gave us the big clue on how to proceed. We had a short discussion but then we settled that a circle was nothing but an infinite sided polygon like a humongous pizza with lots of slices. Tommy was more worried about the pineapple and the ham bits on the Pizza. We just told him to remain quiet, otherwise he would get nothing.

To solve this problem, we decided to start with a 6-slice pizza (Fig. 7.1). First, we examined one slice of the pizza. It started from the center of the pizza at A. We drew a line BC to connect the other two corners of the slice. Instead of the slice, this gave us a triangle ABC. At this time we did not worry about the small amount of the food between the line BC and the edge of the pizza. We argued but decided to agonize with it later.

We drew a vertical dotted line from AD. AD is called the apothem of this hexagon which is somewhat like a radius of a circle. We assumed that AD = r which is approximately the radius of the pizza.

The angle ABC = 360°/6 since the angles for all the slices would add upto 360°. The angle BAD would be half of this and hence 30°. Because ABC is an isosceles triangle, BD = DC. We called BD =L/2 which is half the length of the base BC of the triangle.

Now L/2/r = tan (BAD) = tan 30° = 0.57735.

Area of the triangle ABC = r x L/2 = 0.57735 r^{2}.

But the pizza contained 6 such triangles. So the area of the hexagon in the pizza = 6 x 0.57735 r^{2} = 3.4641 r^{2}. This gave us the value of π to be 3.4641 because area of a circle is 3.4641 r^{2}.

# Double the number of pizza slices

Remember, our premise was that a circle can be thought of a polygon with infinite sides. As a next step towards that, we examined a 12 slice pizza the same way (Fig. 7.2). But Tommy was not happy because that would give him less pizza per slice. So we said that the pizza can be larger but it can be cut into 12 equal slices. Tommy was satisfied. We drew a triangle connecting the center of the pizza and edges of the slices the same way as for the 6 slice pizza. We also drew a similar vertical line from A.

Here BD/AD = L/2/r = tan (360/12 x 2) = tan 15°= 0.26795.

Therefore area of the triangle ABC

= 0.26795 r^{2}.

The area of the dodecagon in the pizza

= 12 x 0.26795 r^{2 }= 3.2154 r^{2}.

This gave us the value of π to be 3.2154 which is somewhat smaller than 3.4641 which was the value we had obtained from the 6-slice pizza.

We noticed that the determination of the value of π from the polygons in the 6 and the 12 slice pizzas followed the same equation. The area of triangle ABC = tan (360°/2n) x r^{2 }and the area of the polygon = n tan (360°/2n) x r^{2} where n is the number of sides.

# 100 slice pizza

We next thought of a 100 slice pizza. Tommy didn’t like the idea again because each slice would be too small and he would remain hungry. So, we told him not to worry because this pizza would be very large like the size our class room.

With the same equation, the area of a 100-sided polygon in the pizza would be:

100 x tan (360°/200) x r^{2} = 100 x 0.03142626604 r^{2} = 3.142626604 r^{2}. Again, this value was slightly smaller than the two values we had obtained before.

For a 10,000 sided polygon, the area would be 10000 x 0.00031415927 r^{2 }= 3.1415927 r^{2}. Here are the values we got for the areas of the different pizzas: 3.4641 r^{2 }(6 slices), 3.2154 r^{2} (12 slices), 3.142626604 r^{2} (100 slices), 3.1415927 r^{2} (10000 slices). We could have kept going but the 20 minutes were up.

Because of the area of Area of a circle = r^{2 }and the area of a 10,000 polygon is 3.1415927 r^{2}, we came up wuth π = 3.1415927 approximately.

# Mr Power was thrilled with the answer.

He told the class that this was brilliant but not the only way. He continued saying that although they had started learning Trig using right angle triangles, they should now know that Trig can be used for any triangles and polygons. They should also explore how Trig can also be used for any quadrilaterals and polyhedrons. He gave them several problems from the Trig book as their homework.

Johnny, by now. liked the amazing things he could do with Trig because of the help of the girlfriend he loved. He wrote the final exam with confidence. Indeed, he obtained 99% total in the course. This mark helped him keep the bike because his overall average from all the courses was well above 85 %. Sara told this story to her Nana (grandma) whom she loved very much. Nana said, “I am not surprised because you are Sarasvati – the goddess of knowledge. Sara is only your nickname.”

*Challenge*

*Challenge:* Johnny enters a bike run. Sara asks him, “How long is the run?’ Johnny, “I don’t know but based on this map, from the starting point we go 20 km on a road going West, after which the North – East path takes an angle of 70°. We go for 18 kilometers on this road and then turn onto another road East with an angle of 110° and we go for 15 kilometers. After that we take a road that brings us back straight to our starting point.” Sara says, “So, the route is a trapezoid.” How long did you figure out the bike run to be?

* Solution: *We draw a quadrangle ABCD (Fig. 10.6). Given: AB = 20 km, BC = 18 km, CD = 15 km, angle ABC = 70°, angle BCD = 110°. Since the sum of the angles ABC and BCD = 180°, the lines AB and CD must be parallel thus making the quadrangle a trapezoid. We need the sum of all the four sides to get the distance for bike route. The lengths of AB, BC and CD are given, but we need to determine the length of DA to solve this problem.

We draw two vertical lines: CE from C to AB and AF from A to CD.

In the triangle EBC, angle EBC = angle ABC = 70°.

Therefore, CE/BC = sin 70° or CE = BC sin 70° = 18 x 0.9397 = 16.9146 km.

BE/BC = cos 70° or BE = BC cos 70° = 18 x 0.342 = 6.156 km.

AE = AB – BE = 20 – 6.156 =13.844 km.

Because AB and CD are parallel, CE and AF being perpendiculars to them must also be parallel.

Therefore AF = CE = 16.9146 km.

and CF = EA = 13.844 km.

Because we are given that CD = 15 km, FD =15 – 13.844 = 1.156 km.

Because AF was drawn as a line perpendicular to CD, the triangle AFD is a right angle triangle. From the Pythagoras theorem

DA^{2} = AF^{2 }+ FD^{2 }= 13.844^{2} + 1.156^{2 }= 192.993 or DA = 13.892 km.

Therefore, the total distance of the bike route = 20 + 18 +15 + 13.892 = 66.892 km.